# What i do wrong(find the current)

1. Dec 2, 2011

### omni

in the picture you can see the electric circuit.
i asked to find the current in R3.

i mark 4 circle (a b c d)
and i got the 4 equations (shown in blue color).

in equation C, i not sure if is equal to 100 or to 0, but i tried with both of them.

to get the 4 variable.
but for some reason i am not get the correct answer is should be 6.666A.
what i do wrong?

thanks.

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2. Dec 2, 2011

### Staff: Mentor

For loop C there is a 100V source there, so your equation should = 100.

Otherwise the equations appear to be okay. What values do you get for the mesh currents?

3. Dec 3, 2011

### omni

well the values i get for the mesh currents is: a=-10/3, b=-70/3, c=10/3, d=-100/3
so i know i need to change direction for a,and b circles.

the current in R3 are affected from the currents a,b,c ? i get the correct answer that should be 6.666A, only if i take a and c and addition them together.

what do you think?
thanks.

4. Dec 3, 2011

### Staff: Mentor

The currents that you found are correct. They correspond to the mesh currents as you've drawn them (meaning that negative values imply that the mesh current is actually flowing in the opposite direction to the assumed direction. But that's fine! Leave the circuit drawing alone and just keep the negative values -- they are fine together.

When you wrote the mesh current equations you had terms in each of them that represented the currents from other meshes that flowed in shared components. That's how, for example, terms -10b, -10c, and 0d appeared in the equation for the first mesh. What this means is that TWO mesh currents can flow in shared components, and the net current in that component is then the sum of those two currents.

For R3, mesh current A flows down through it and mesh current C flows up through it. So the total flowing in the downward direction (if that's what you want) is I = a - c. Your values of a and c are -10/3 and 10/3 amps respectively. So I = -10/3 - 10/3 = -20/3. The negative sign indicates that the "real" net current flows upwards through R3.

5. Dec 3, 2011

### omni

again you make it clear to understand.

i really appreciate your help :)

6. Dec 3, 2011

### Staff: Mentor

And I'm happy to help! Cheers!